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x-n-y-n-n-N-




Question Number 194786 by tri26112004 last updated on 15/Jul/23
x^n +y^n =¿ (n∈N^∗ )
$${x}^{{n}} +{y}^{{n}} =¿\:\left({n}\in{N}^{\ast} \right) \\ $$
Answered by Frix last updated on 15/Jul/23
Let y=px  x^n +y^n =x^n (p^n +1)=x^n Π_(k=1) ^n (p−e^((iπ(1+2k))/n) )  Let p=(y/x)  x^n Π_(k=1) ^n (p−e^((iπ(1+2k))/n) )=x^n Π_(k=1) ^n ((y/x)−e^((iπ(1+2k))/n) )
$$\mathrm{Let}\:{y}={px} \\ $$$${x}^{{n}} +{y}^{{n}} ={x}^{{n}} \left({p}^{{n}} +\mathrm{1}\right)={x}^{{n}} \underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left({p}−\mathrm{e}^{\frac{\mathrm{i}\pi\left(\mathrm{1}+\mathrm{2}{k}\right)}{{n}}} \right) \\ $$$$\mathrm{Let}\:{p}=\frac{{y}}{{x}} \\ $$$${x}^{{n}} \underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left({p}−\mathrm{e}^{\frac{\mathrm{i}\pi\left(\mathrm{1}+\mathrm{2}{k}\right)}{{n}}} \right)={x}^{{n}} \underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\frac{{y}}{{x}}−\mathrm{e}^{\frac{\mathrm{i}\pi\left(\mathrm{1}+\mathrm{2}{k}\right)}{{n}}} \right) \\ $$

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